Question

When 70.4 g of benzamide (C,H,NO) are dissolved in 850. g of a certain mystery liquid X, the freezing point of the solution is 2.7 C lower than the freezing point of pure X. On the other hand, when 70.4 g of ammonium chloride (NH CI) are dissolved in the same mass of X, the freezing point of the solution is 9.9 °C lower than the freezing point of pure X. Calculate the van't Hoff factor for ammonium chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits.

Answer 1

Answer:

1.60 is the van't Hoff factor for ammonium chloride in X.

Explanation:

$\Delta T_f=iK_f\times m$

$Delta T_f=K_f\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}$...(1)

where,

$\Delta T_f$ =Elevation in boiling point =

i = van't Hoff factor

$K_f$ = Freezing point constant

m = molality

1) When 70.4 g of benzamide  are dissolved in 850. g of a certain mystery liquid X.

Mass of benzamide = 70.4 g

Molar mass of benzamide = 121 g/mol

i = 1 (organic molecule)

Mass of liquid X = 850 g = 0.850 kg

$K_f$ = Freezing point constant of liquid X= ?

$\Delta T_f=2.7^oC$

Putting all value in a (1):

$2.7^oC=K_f\times \frac{70.4 g}{121 g/mol\times 0.850 kg}$

$K_f=3.944 ^oC kg/mol$

2) When 70.4 g of ammonium chloride are dissolved in 850. g of a certain mystery liquid X.

Mass of ammonium chloride= 70.4 g

Molar mass of ammonium chloride = 53.5 g/mol

i = ?  (ionic molecule)

Mass of liquid X = 850 g = 0.850 kg

$K_f=3.944 ^oC kg/mol$

$\Delta T_f=9.9^oC$

Putting all value in a (1):

$9.9^oC=i\times 3.944^oC kg/mol\times \frac{70.4 g}{53.5 g/mol\times 0.850 kg}$

i = 1.6011 ≈ 1.60

1.60 is the van't Hoff factor for ammonium chloride in X.